Question: Rewrite the equation by completing the square. $x^{2}-6x-16 = 0$ $(x + $
Answer: Begin by moving the constant term to the right side of the equation. $x^2 - 6x = 16$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. Since the coefficient of our $x$ term is $-6$, half of it would be $-3$, and squaring it gives us ${9}$. $x^2 - 6x { + 9} = 16 { + 9}$ We can now rewrite the left side of the equation as a squared term. $( x - 3 )^2 = 25$ This is equivalent to $(x+{-3})^2=25$